# Separable polynomial

In mathematics, two slightly different notions of separable polynomial are used, by different authors.

According to the most common one, a polynomial P(X) over a given field K is separable if its roots are distinct in an algebraic closure of K, that is the number of its distinct roots is equal to its degree.[1] In the context of polynomial factorization, such a separable polynomial is also called a square-free polynomial.

For the second definition, P(X) is separable if each of its irreducible factors in K[X] have distinct roots in an algebraic closure of K, that is, if each of its irreducible factors is square-free.[2] In this definition, separability depends on the field K, as for example an irreducible polynomial P which is not separable becomes separable over the splitting field of K. Also, for this definition every polynomial over a perfect field is separable, which includes in particular all fields of characteristic 0, and all finite fields.

Both definitions coincide in case P(X) is irreducible over K, which is the case used to define the notion of a separable extension of K.

In the remainder of this article, only the first definition is used.

A polynomial P(X) is separable if and only if it is coprime to its formal derivative P′(X).

## Separable field extensions

Separable polynomials are used to define separable extensions: A field extension $K \subset L$ is a separable extension if and only if for every $\alpha\in L$, which is algebraic over K, the minimal polynomial of $\alpha$ over K is a separable polynomial.

Inseparable extensions (that is extensions which are not separable) may occur only in characteristic p.

The criterion above leads to the quick conclusion that if P is irreducible and not separable, then P′(X)=0. Thus we must have

P(X) = Q(Xp)

for some polynomial Q over K, where the prime number p is the characteristic.

With this clue we can construct an example:

P(X) = XpT

with K the field of rational functions in the indeterminate T over the finite field with p elements. Here one can prove directly that P(X) is irreducible, and not separable. This is actually a typical example of why inseparability matters; in geometric terms P represents the mapping on the projective line over the finite field, taking co-ordinates to their pth power. Such mappings are fundamental to the algebraic geometry of finite fields. Put another way, there are coverings in that setting that cannot be 'seen' by Galois theory. (See radical morphism for a higher-level discussion.)

If L is the field extension

K(T1/p),

in other words the splitting field of P, then L/K is an example of a purely inseparable field extension. It is of degree p, but has no automorphism fixing K, other than the identity, because T1/p is the unique root of P. This shows directly that Galois theory must here break down. A field such that there are no such extensions is called perfect. That finite fields are perfect follows a posteriori from their known structure.

One can show that the tensor product of fields of L with itself over K for this example has nilpotent elements that are non-zero. This is another manifestation of inseparability: that is, the tensor product operation on fields need not produce a ring that is a product of fields (so, not a commutative semisimple ring).

If P(x) is separable, and its roots form a group (a subgroup of the field K), then P(x) is an additive polynomial.

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## Applications in Galois theory

Separable polynomials occur frequently in Galois theory.

For example, let P be an irreducible polynomial with integer coefficients and p be a prime number which does not divides the leading coefficient of P. Let Q be the polynomial over the finite field with p elements, which is obtained by reducing modulo p the coefficients of P. Then, if Q is separable (which is the case for every p but a finite number) then the degrees of the irreducible factors of Q are the lengths of the cycles of some permutation of the Galois group of P.

Another example: P being as above, a resolvent R for a group G is a polynomial whose coefficients are polynomials in the coefficients of p, which provides some information on the Galois group of P. More precisely, if R is separable and has a rational root then the Galois group of P is contained in G. For example, if D is the discriminant of P then $X^2-D$ is a resolvent for the alternating group. This resolvent is always separable if P is irreducible, but most resolvents are not always separable.

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## References

1. ^ S. Lang, Algebra, p. 178
2. ^ N. Jacobson, Basic Algebra I, p. 233
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